Find the x-intercept(s) and the coordinates of the vertex for the parabola . If there is more than one -intercept, separate them with commas.

Question

y = x² + 4x - 5

Answer 1

You could do something more elaborate and change the equation to standard form by completing the square.

The x-term is 4. Divide 4 in half (which equals 2) then square 2. 2^2 equals 4 (it's just coincidence that it's the same number in this case). now add and subtract 4 from the original equation.

y = x^2 + 4x - 5

y = x^2 + 4x + 4 - 4 - 5

now re-group

y = x^2 + 4x + 4 [-4-5]

simplify

y = x^2 + 4x + 4 - 9

now factor the first 3 terms

y = (x+2)(x+2) - 9

put into parabola form

y = (x+2)^2 - 9

now the problem is easy!

the vertex is at (2, -9)

the parabola crosses the x-axis when y=0; so set the equation equal to zero and solve for x.

0 = (x+2)^2 - 9

add 9 to both sides

9 = (x+2)^2

take the square root of both sides

3 = (x+2) OR -3 = (x+2)

subtract 2 from both sides.

1 = x. OR. -5 = x

Voila!

Answer 2

If the parabola is Ax² + Bx + C = f(x)

The vertex occurs where x=-B/2A; in this case where x = -2 and the parabola crosses the x-axis at x = 1 and x = -5. These values were obtained by factoring (x).